Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int limit = 1000000;
double Y, a, b;
int n;
double l[limit], r[limit], len[limit], sumlen[limit];
double x, y;
double getpos(double despx) { return x + (despx - x) / y * (y + Y); }
void write(double sol) { printf("%.12f\n", sol); }
int main() {
  int inY, ina, inb;
  scanf("%d%d%d", &inY, &ina, &inb);
  Y = inY;
  a = ina;
  b = inb;
  Y *= -1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int inl, inr;
    scanf("%d%d", &inl, &inr);
    l[i] = inl;
    r[i] = inr;
    len[i] = r[i] - l[i];
    sumlen[i] = sumlen[i - 1] + len[i];
  }
  int cases;
  scanf("%d", &cases);
  for (int cas = 0; cas < cases; cas++) {
    int inx, iny;
    scanf("%d%d", &inx, &iny);
    x = inx;
    y = iny;
    int leftmost;
    double sol = 0;
    {
      int inf = 0;
      int sup = n + 1;
      while (inf + 1 < sup) {
        int med = (inf + sup) / 2;
        if (a < getpos(r[med]))
          sup = med;
        else
          inf = med;
      }
      leftmost = sup;
      if (leftmost == n + 1 or b < getpos(l[leftmost])) {
        write(0.0);
        continue;
      }
      if (getpos(l[leftmost]) < a) {
        if (b < getpos(r[leftmost])) {
          write(b - a);
          continue;
        } else {
          sol += getpos(r[leftmost]) - a;
          leftmost++;
        }
      } else {
        if (b < getpos(r[leftmost])) {
          write(b - getpos(l[leftmost]));
          continue;
        }
      }
    }
    int rightmost;
    {
      int inf = 0;
      int sup = n + 1;
      while (inf + 1 < sup) {
        int med = (inf + sup) / 2;
        if (getpos(l[med]) < b)
          inf = med;
        else
          sup = med;
      }
      rightmost = inf;
    }
    if (rightmost == 0 or getpos(r[rightmost]) < a) {
      write(0.0);
      continue;
    }
    if (b < getpos(r[rightmost])) {
      if (getpos(l[rightmost]) < a) {
        write(b - a);
        continue;
      } else {
        sol += b - getpos(l[rightmost]);
        rightmost--;
      }
    } else {
      if (getpos(l[rightmost]) < a) {
        write(getpos(r[rightmost]) - a);
        continue;
      }
    }
    if (leftmost <= rightmost)
      sol += getpos(sumlen[rightmost] - sumlen[leftmost - 1]) - getpos(0);
    write(sol);
  }
}